Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $n = \dfrac{-3y - 18}{5y - 35} \times \dfrac{y^2 - 13y + 42}{-4y^2 + 4y + 168} $
Solution: First factor out any common factors. $n = \dfrac{-3(y + 6)}{5(y - 7)} \times \dfrac{y^2 - 13y + 42}{-4(y^2 - y - 42)} $ Then factor the quadratic expressions. $n = \dfrac {-3(y + 6)} {5(y - 7)} \times \dfrac {(y - 7)(y - 6)} {-4(y + 6)(y - 7)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {-3(y + 6) \times (y - 7)(y - 6) } {5(y - 7) \times -4(y + 6)(y - 7) } $ $n = \dfrac {-3(y - 7)(y - 6)(y + 6)} {-20(y + 6)(y - 7)(y - 7)} $ Notice that $(y + 6)$ and $(y - 7)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {-3(y - 7)(y - 6)\cancel{(y + 6)}} {-20\cancel{(y + 6)}(y - 7)(y - 7)} $ We are dividing by $y + 6$ , so $y + 6 \neq 0$ Therefore, $y \neq -6$ $n = \dfrac {-3\cancel{(y - 7)}(y - 6)\cancel{(y + 6)}} {-20\cancel{(y + 6)}(y - 7)\cancel{(y - 7)}} $ We are dividing by $y - 7$ , so $y - 7 \neq 0$ Therefore, $y \neq 7$ $n = \dfrac {-3(y - 6)} {-20(y - 7)} $ $ n = \dfrac{3(y - 6)}{20(y - 7)}; y \neq -6; y \neq 7 $